Problem: A few families took a trip to an amusement park together. Tickets cost $$5.50$ each for adults and $$3.00$ each for kids, and the group paid $$26.00$ in total. There were $3$ fewer adults than kids in the group. Find the number of adults and kids on the trip.
Let $x$ equal the number of adults and $y$ equal the number of kids. The system of equations is then: ${5.5x+3y = 26}$ ${x = y-3}$ Solve for $x$ and $y$ using substitution. Since $x$ has already been solved for, substitute ${y-3}$ for $x$ in the first equation. ${5.5}{(y-3)}{+ 3y = 26}$ Simplify and solve for $y$ $ 5.5y-16.5 + 3y = 26 $ $ 8.5y-16.5 = 26 $ $ 8.5y = 42.5 $ $ y = \dfrac{42.5}{8.5} $ ${y = 5}$ Now that you know ${y = 5}$ , plug it back into ${x = y-3}$ to find $x$ ${x = }{(5)}{ - 3}$ ${x = 2}$ You can also plug ${y = 5}$ into ${5.5x+3y = 26}$ and get the same answer for $x$ ${5.5x + 3}{(5)}{= 26}$ ${x = 2}$ There were $2$ adults and $5$ kids.